## Chemistry## Equilibrium ConstantsAll equilibria have different ratios of products to reactants. As an example, let's consider a reaction in which the forward reaction takes place quickly and the reverse reaction is slow. When the system reaches equilibrium, a vast majority of the chemicals in the beaker will be products (we usually say that the "equilibrium favors the products"). On the other hand, if the forward reaction is slow and the reverse reaction is fast, the equilibrium favors the reactants. It probably won't be a big surprise to you to find that scientists have figured out a way to quantify the position of an equilibrium. This was done by the "law of mass action." What the law of mass action says is simple. Let's say that we have a reaction taking place in solution with the following equation: - aA + bB ⇔ cC + dD
The equilibrium condition can be expressed using the following equation: *K*_{eq}=^{[C]c [D]d}⁄_{[A]a[B]b}
## The Mole SaysWe can relate the equilibrium constant of a chemical equilibrium to the rates of the forward and reverse reactions. For example, consider the process A ⇔ B. The rate of the forward reaction, A → B, is k where K The equilibrium constant is important because it gives us an idea of where the equilibrium lies. The larger the equilibrium constant, the further the equilibrium lies toward the products. For example, an equilibrium constant of 1.0 × 10 K
- C
_{2}H_{3}O_{2}H_{(aq)}+ C_{2}H_{3}O_{2}^{-}_{(aq)}
Write the expression for the equilibrium constant, and determine what the value of K
To figure out the expression for the chemical equilibrium, use the equation we learned for finding K *K*_{eq}=^{[C2H3O2-][H+]}⁄_{[C2H3O2H]}
## Bad ReactionsBe careful not to mistake the equilibrium constant K To determine the value of K *K*_{eq}=^{(1.0×10-4M)(1.0×10-4M)}⁄_{5.6×10-4M)}- K
_{eq}= 1.8 × 10^{-5}
Once we have an equilibrium constant, we can use it to figure out what the equilibrium concentrations of the products will be given an initial concentration of the reactants. Let's see another example:
*K*=_{eq}^{[B][C]}⁄_{[A]}
Our next step is to figure out what the concentrations of each species will be at equilibrium. We do this by setting up a chart that shows the initial concentrations of all species, how the concentrations of each will change, and what the final concentrations of each species will be. For this process, the chart is given next (don't panic, we'll explain how we got all these values in a minute):
Okay, let's talk about where these came from: - The initial concentration of A is defined by the problem as 1.00 M. Between the time that the reaction starts and the system reaches equilibrium, some of it will turn into the products. How much? We have no idea, so we'll just say that the change was "-x." As a result, our final concentration will be the initial concentration minus the amount of change, or (1.00 - x) M.
- The initial concentrations of both B and C are zero because neither was initially present. However, from the equation, you can see that every time a molecule of A breaks apart, one molecule each of B and C are formed. As a result, if the concentration of A decreases by x, the concentration of both products must increase by x. At equilibrium, the concentration of both B and C is x.
To figure out what the concentrations of each species are, we plug these values into the expression for finding the equilibrium constant. Since we know that the equilibrium constant is 1.86 × 10 - 1.86×10
^{-6}=^{[x][x]}⁄_{[1.00 - x]}
Now, solving for x won't be a lot of fun because we'll need to use the quadratic equation. Let's face it, we don't really want to memorize the quadratic equation, much less use it. ## You've Got ProblemsProblem 1: Given the reaction H a) The general expression for the equilibrium constant. b) The equilibrium concentration of HI if I start with 2.00 M H Fortunately, there's a shortcut we can use to get around using the quadratic equation when the K - 1.86×10
^{-6}=^{[x][x]}⁄_{[1.00]} - x = 1.36 × 10
^{-3}M
Let's see if this was a good assumption. If the amount of product formed was 1.36 × 10 Our final concentrations of B and C are 1.36 × 10 Excerpted from The Complete Idiot's Guide to Chemistry © 2003 by Ian Guch. All rights reserved including the right
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